Time Domain Method of Moments
Massachusetts Institute of Technology
6.635 lecture notes
1
Introduction
The Method of Moments (MoM) introduced in the previous lecture is widely used for solving
integral equations in the frequency domain. Yet, some attempts have been made recently at the
use of the MoM in the time domain. We shall briefly expose this approach here.
2
Time domain equations
The first step is of course to write Maxwell’s equation and all other relations (constitutive
relations and continuity) in time domain:
∂
B̄(r̄, t) − M̄ (r̄, t) ,
∂t
∇ · B̄(r̄, t) = m(r̄, t) ,
∂
¯ t) ,
D̄(r̄, t) + J(r̄,
∂t
∇ · D̄(r̄, t) = ρ(r̄, t) ,
D̄(r̄, t) = ²Ē(r̄, t) ,
¯ t) + ∂ ρ(r̄, t) = 0 ,
∇ · J(r̄,
∂t
B̄(r̄, t) = µH̄(r̄, t) ,
∂
∇ · M̄ (r̄, t) + m(r̄, t) = 0 .
∂t
∇ × Ē(r̄, t) = −
∇ × H̄(r̄, t) =
(1a)
(1b)
(1c)
(1d)
For the time-domain MoM, it is easier to work with the potentials, and make use of the
well-known “retarded potentials” theory. In view of doing this, we write the definition:
1
∇ × Ā(r̄, t)
µ0
∂
Ē(r̄, t) = −∇φ(r̄, t) − Ā(r̄, t) .
∂t
H̄(r̄, t) =
(2a)
(2b)
Both the vector potential Ā and the scalar potential φ satisfy the wave equation which, in
time-domain domain, writes:
∂2
¯ t) ,
Ā(r̄, t) = −µ0 J(r̄,
∂t2
∂2
ρ(r̄, t)
∇2 φ(r̄, t) − ²0 µ0 2 φ(r̄, t) = −
.
∂t
²0
∇2 Ā(r̄, t) − ²0 µ0
(3a)
(3b)
These potentials are linked by the time-domain Lorentz gauge:
∇ · Ā(r̄, t) + ²0 µ0
1
∂
φ(r̄, t) = 0 .
∂t
(4)
2
Section 2. Time domain equations
We can defined also a time-domain Green’s function which satisfies the time-domain scalar
equation:
1 ∂2
(∇2 − 2 2 )g(r̄, r̄ 0 , t, t0 ) = −δ(r̄ − r̄ 0 ) δ(t − t0 ) ,
(5)
c ∂t
which solution is (in free-space):

|r̄−r̄ 0 |
 1
0
0
c ) t>t,
4π|r̄−r̄ 0 | δ(t − t −
0
0
(6)
g(r̄, r̄ , t, t ) =
0
t < t0 .
From this, the solution to the wave equation for Ā and φ can be written as:
Z
Z ∞
Z
¯ 0 , t − R/c)
J(r̄
0
0¯ 0 0
0
0
dv
dv 0
Ā(r̄, t) =µ0
dt J(r̄ , t ) g(r̄, r̄ , t, t ) = µ0
,
4πR
V
V
−∞
φ(r̄, t) =
Z
dv 0
V
ρ(r̄, t − R/c)
,
4π²0 R
(7a)
(7b)
where R = |r̄ − r̄ 0 |. These wave equations are known as the time retarded potentials, and
essentially say that the potential (either Ā or φ) can be calculated at a given point in space r̄
and given time t from all previous times.
From these equations, we can calculate the space-time electromagnetic fields:
Z
¯ 0, τ )
1
J(r̄
H̄(r̄, t) =
dv 0 ∇ ×
,
τ = t − R/c ,
(8a)
4π
R
Z
Z
¯ 0, τ )
ρ(r̄, τ ) µ0
1
∂ J(r̄
0
dv ∇
Ē(r̄, t) = −
−
dv 0
.
(8b)
4π²0 V
R
4π
∂t R
Let us continue with the electric field first:
·
¸
Z
Z
1
1
µ0
1 ∂ ¯ 0
1
dv 0
∇ρ(r̄0 , τ ) + ρ(r̄ 0 , τ )∇
−
Ē(r̄, t) = −
dv 0
J(r̄ , τ ) .
4π²0 V
R
R
4π V
R ∂t
(9)
At this point, we need to use the following relations:
R̄
,
R
1
R̄
∇ =− 3,
R
R
∂
1
∂
1 R̄ ∂
∇ρ(r̄0 , τ ) =
ρ(r̄0 , τ )∇τ = − ∇R ρ(r̄0 , τ ) = −
ρ(r̄, τ ) .
∂τ
c
∂τ
c R ∂τ
∇R =
We can therefore continue with the electric field as:
·
¸
Z
Z
1 ∂ ¯ 0
µ0
1
R̄
0 1 R̄ ∂
0
0
dv 0
dv
J(r̄ , τ )
Ē(r̄, t) =
ρ(r̄ , τ ) + 3 ρ(r̄ , τ ) −
2
4π²0
c R ∂τ
R
4π
R ∂τ
¸
·
Z
Z
R̄
1
µ0
1 ∂ ¯ 0
1
1 ∂
J(r̄ , τ ) .
ρ(r̄0 , τ ) + ρ(r̄0 , τ ) 2 −
dv 0
=
dv 0
4π²0
c ∂τ
R
R
4π
R ∂τ
(10a)
(10b)
(10c)
(11)
3
We can perform the same type of calculations for the magnetic field using the relation
¯ 0 , τ ) = − 1 R̄ × ∂ J(r̄
¯ 0, τ ) .
∇ × J(r̄
c R ∂τ
(12)
We get:
H̄(r̄, t) =
1
4π
Z
·
¸
R̄
1 R̄
∂ ¯ 0
¯ 0, τ ) .
dv 0 −
×
J(r̄
,
τ
)
−
×
J(r̄
c R2 ∂τ
R3
(13)
Upon gathering the expressions for the electric and magnetic field, we eventually get:
½·
¾
¸
Z
1
1 ∂
1
R̄
µ0 ∂ ¯ 0
0
0
0
Ē(r̄, t) =
dv
J(r̄ , τ )
ρ(r̄ , τ ) + ρ(r̄ , τ )
−
4π
c ∂τ
R
²0 R 2
R ∂τ
·
¸
Z
R̄
1
1 ¯ 0
1 ∂ ¯ 0
dv 0
J(r̄ , τ ) + J(r̄
, τ) × 2 .
H̄(r̄, t) =
4π
c ∂τ
R
R
(14a)
(14b)
Upon using the boundary conditions for the electric and magnetic field, we construct the
integral equations in a standard way:
• EFIE: n̂ × (Ē i + Ē scat ) = 0 on PEC surface
1
n̂ ×
⇒ n̂ × Ē (r̄, t) +
4π
i
Z
ds0 [. . .]
(15)
• MFIE: n̂ × (H̄ i + H̄ scat ) = J¯s .
As we have seen before (in a previous class), this integral equation is expressed in terms
of the principal value of the integral with a 1/2 additional factor. Thus:
Z
1¯
1
J(r̄, t) = n̂ × H̄ i (r̄, t) +
n̂ × P V
ds0 [. . .]
(16)
2
4π
For the sake of comparison, we can write the MFIE in the frequency domain and in the time
domain:
Z
i
¯
¯ 0 ) × ∇0 g(r̄, r̄0 )
J(r̄) = 2n̂ × H̄ (r̄) + 2n̂ × P V
ds0 J(r̄
r̄ ∈ S ,
(17a)
·
¸
Z
1 ∂ ¯ 0
1 ¯ 0
R̄
¯ t) = 2n̂ × H̄ i (r̄) + 1 n̂ × P V
ds0
(17b)
J(r̄ , τ ) + J(r̄
, τ) × 2 .
J(r̄,
2π
c ∂τ
R
R
Note that in the principal value, we essentially exclude the part for which R = 0. Since
τ = t − R/c and R 6= 0, we always have that τ < t. The time domain equations therefore
state that the current at location r̄ and time t is equal to a known term 2n̂ × H̄ i (r̄, t) plus a
¯ This is the basis for solving the time domain
term (integral) known from the past history of J.
integral equation by iterative methods, the most well-known one being the marching-on-in-time.
4
Section 3. The marching-on-in-time technique
3
The marching-on-in-time technique
3.1
General equations
The integral equation can often be cast in the following form:
Z
Z τ
i
0
¯
¯
¯ 0 , t0 ) .
J(r̄, t) = J (r̄, t) +
dv
dt0 K(r̄, r̄ 0 , t − t0 ) · J(r̄
(18) eq.10
0
S
eq.10
¯ 0 , t0 ) has not yet been set to satisfy
Note that we have a time integral also as in Eq. (18), J(r̄
any causality condition. Hence, we must then impose J¯i (r̄, t) = 0 for t < 0, r̄ ∈ S.
In order to apply the MoM, we discretize the current both in space and in time:
¯ 0 , t0 ) =
J(r̄
N
M
X
X
m0 =1
J¯p (m0 , n0 ) Ps (r̄0 − r̄m0 ) Pt (t0 − tn0 ) ,
(19)
n0 =0
where P denotes the simple pulse function. In addition, we also apply point-matching, which
means that we take the following testing functions:
Wmn (r̄m , tn ) = δ(r̄ − r̄m ) δ(t − n∆t) = δ(r̄ − r̄m ) δ(t − tn ) ,
(20)
0 |. The method is best illustrated on the
where we take ∆t = min{Rmm0 /c}, Rmm0 = |r̄m − r̄m
following example.
3.2
Example
Let us consider a 1D example governed by the following integral equation:
g(x, t) =
Z
x0
K(x, x0 ) f (x0 , τ )dx0 ,
x ∈ [−x0 , x0 ],
τ = τ (x, x0 , t) = t −
−x0
|x − x0 |
.
c
(21)
Let us chose the following expansion for f :
f (x0 , τ ) '
N X
J
X
i0 =1
ai0 j 0 Pi0 j 0 (x0 , τ ) ,
(22)
j 0 =1
where the pulse basis functions are defined as

1 for x0 ∈ [x 0 − dx , x 0 +
i
i
2
P i0 j 0 =
0 elsewhere.
dx
2 ]
and t ∈ [tj 0 −
dt
0
2 , tj
+
dt
2]
(23)
Note that we use the definitions: xi0 = i0 dx , tj 0 = j 0 dt , and dx = cdt . In order to apply point
matching, we take the following testing functions:
wij (x, t) = δ(x − xi ) δ(t − tj ) .
(24)
5
Upon expanding and testing, we get:
g(xi , tj ) = gij =
=
Z
x0
0
K(xi , x )
−x0
J
N X
X
ai0 j 0 Pi0 j 0 (x0 , τ )δ(t − tj )
i0 =1 j 0 =1
N Z (i0 + 1 )dx
X
2
0 1
i0 =1 (i − 2 )dx
dx
0
J
X
ai0 j 0 Pi0 j 0 (x0 , τ )K(idx , x0 )δ(t − tj ) .
(25)
j 0 =1
Coming back to the definition of τ , we write (with the test and the expansion):
τ = tj −
dx
|xi − xi0 |
= jdt − |i − i0 | = (j − |i − i0 |)dt ,
c
c
(26)
such that the coefficient ai0 j 0 becomes ai0 ,j−|i−i0 | . The integral equation becomes:
gij =
N
X
ai0 ,j−|i−i0 |
i0 =1
We can define the term
Zii0 =
Z
Z
(i0 + 12 )dx
(i0 − 21 )dx
(i0 + 21 )dx
(i0 − 21 )dx
dx0 K(idx , x0 ) .
dx0 K(idx , x0 )
(27)
(28)
and rewrite the previous system as
gij =
N
X
Zii0 ai0 ,j−|i−i0 |
i0 =1
= Zii aij + Zi,i−1 ai−1,j−1 + Zi,i−2 ai−2,j−2 + . . . + Zi1 a1,j−i+1
+ Zi,i+1 ai+1,j−1 + Zi,i+2 ai+2,j−2 + . . . + Z1,N aN,j−N +i .
(29)
In this equation, only the first term involves time step j, all the others terms being at j − 1,
j − 2, . . . Therefore, we can solve for aij :
¸
·
N
X
1
aij =
Zii0 ai0 ,j−|i−i0 | .
gij −
Zii
0
(30)
i =1
i0 6=i
The value of all aik are known for k < j, so that aij is completely specified in closed form by
those and the present value of gij . This process is known as a 1D march-on-in-time approach.
Time-domain MoM is nowadays in its early stage and, although it has been successfully
applied to various simple situations, still suffers from numerical instabilities. More work is in
progress...